The definition of submanifold in my class is as follows.
Let $d$ and $n$ natural numbers, such that $0 \leq d \leq n$, let $l \in \mathbb{N} \cup \{\infty \} $, and let $X$ a normed vector space of dimension $n$. A subset $M \subset X$ is a $d$-dimensional $C^l$-submanifold of $X$ if it satisfies:
For all $p \in M$ exist a open subset $U \subset X$ whit $p \in U$, a open $V \subset \mathbb{R}^n$, and a $C^l$-diffeomorphism $\phi:U \rightarrow V$ with $$\phi(U \cap M)=V \cap (\mathbb{R}^d \times\{0\}).$$
I have to prove that the set $S$ is not a submanifold of $\mathbb{R}^2$, where $S:=\{(x,y) \in \mathbb{R}^2~|~y\neq 0 \Rightarrow 1/y \in \mathbb{N} \}$.
My problem is that I believe I have demonstrated that $S$ is a submanifold 1-dimensional of $\mathbb{R}^2$. Then $S$ can be a submanifold of $\mathbb{R}^2$?
Best Answer
Because the definition is not clear I will do two cases:
Proof: Take $p\in S$. Write $p=(x,1/n)$ for suitable $x$ and $n$. Define $U=\Bbb R\times \left(1/n-\epsilon,1/n+\epsilon\right)$ with $\epsilon$ such that $$1/(n+1)<1/n-\epsilon<1/n<1/n+\epsilon<1/(n-1).$$ Finally define $V=\Bbb R\times (-\epsilon, \epsilon)$ and $\phi:U\to V$ by $\phi (a,b)=(a,b-1/n)$. Then $\phi$ is a diffeomorphism and satisfies $$\phi (U\cap S)=\phi(\Bbb R\times \{1/n\})=\Bbb R\times \{0\}=V\cap (\Bbb R\times \{0\}).$$
Proof: Assume by contradiction that $S$ is a submanifold of $\Bbb R^2$. Because $(0,0)\in S$ there exist open sets $U$ and $V$ and a diffeomorphism $\phi:U \rightarrow V$ with $\phi(U \cap S)=V \cap (\mathbb{R}^d \times\{0\})$, such that $(0,0)\in U$ and for instance $\phi(0,0)=(0,0)$. You can decrease $V$ to any open set $V^\prime \subset V$ containing $(0,0)$, and if you take $U^\prime=\phi^{-1}(V^\prime)$ you get that $\phi:U^\prime \rightarrow V^\prime$ is a diffeomorphism with $\phi(U^\prime \cap S)=V^\prime \cap (\mathbb{R}^d \times\{0\})$. In this way you can choose $V$ to be of the form $V=(-\epsilon,\epsilon)\times (-\epsilon,\epsilon)$. Finally $\phi$ restrict to a diffeomorphism $$\phi : U-S\longrightarrow V-\mathbb{R}^d \times\{0\}$$ The problem with this statement is that $V-\mathbb{R}^d \times\{0\}$ has at most two path components, but $U-S$ has infinitely many of them. This is because there are infinitely many $n\geq 1$ such that there is some $x\in \Bbb R$ satisfying $(x,1/n)\in U$ and if $(x,1/n)$ and $(x^\prime,1/m)$ can be joined by a path in $U-S$ they can be joined by a path in $\Bbb R^2-S$ which is equivalent to $m=n$ (look at the second coordinate of such a path and use the intermediate value theorem).
I hope this helps!